I understand that $H(n)=2$ is not true, because the base case for $H(1) \neq 2$, but I don't understand why we would ever think $H(n)=2$, since $H(2n+1) \neq 2$. Any help? This is my first post, so apologies if my question is not appropriate.

Why would we assume $H(n)=2$? I'm having trouble wrapping my head around that.
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dylamDec 11 '12 at 22:43

That's how proofs by induction go. You prove a base case, then you show that if it holds true in some situation it then must also hold true in the next situation. $H(n)=2$ is the induction hypothesis.
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Gerry MyersonDec 11 '12 at 22:49

I get the induction part, but just to be sure: So it's not because $2n$ is a multiple of $n$? I was thinking because $H(2n)=2$ and $2n$ is a multiple of $n$, then we can assume $H(n)=2$.
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dylamDec 11 '12 at 22:57

It has nothing to do with $2n$ being a multiple of $2$. We're trying to prove $H(n)=2$, by induction. The base case, $H(1)=2$, is false; but if we are careless enough to forget to check that base case, then we would proceed to take $H(n)=2$ as the induction hypothesis, and we would think we had constructed a legitimate proof.
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Gerry MyersonDec 11 '12 at 23:31

Ah, as the question says, "Therefore it seems possible to prove that H(n)=2 for all n, by induction on n." Thanks for your help.
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dylamDec 11 '12 at 23:34